First slide

Simple hormonic motion

Question

In fig. (6), the spring has a force constant k The pulley

s light and smooth, the spring and string are light. The suspended block has a mass m. If the block is slightly displaced from its equilibrium position and then released, the period of its vertical oscillation is

Moderate

A

$T = 2 π \sqrt{(\frac{m}{k})}$
B

$T = 4 π \sqrt{(\frac{m}{k})}$
C

$T = 2 π \sqrt{(\frac{k}{m})}$
D

$T = 4 π \sqrt{(\frac{k}{m})}$

Solution

see fig (12)
Let the tension in spring at equilibrium position for block be $T_{0} . Then mg = T_{0}$ From force diagram of pulley
$2 T_{0} = {ky}_{0} or 2 mg = {ky}_{0} When the block is displaced through a small distanceyin downward direction from equilibrium position, let the tension in string be T . The stretching of string will be (y / 2) . Therefore and 2 T = k (y_{0} + \frac{y_{0}}{2}) = {ky}_{0} + \frac{{ky}_{0}}{2} From eqs . (1) and 2), we get 2 T = 2 mg + \frac{{ky}_{0}}{2} or T = mg + (\frac{{ky}_{0}}{4}) Now mg - (mg + \frac{{ky}_{0}}{4}) = ma or ma = - \frac{{ky}_{0}}{4} or a = - (\frac{k}{4 m}) y_{0} therofore \frac{d^{2} y_{0}}{{dt}^{2}} = - (\frac{k}{4 m}) y_{0} or \frac{d^{2} y_{0}}{{dt}^{2}} + (\frac{k}{4 m}) y_{0} = 0 T = \frac{2 π}{ω} = 2 π \sqrt{(\frac{4 m}{R})}$

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